Find $(-1)^{-10} + (-1)^{-9} + (-1)^{-8} + \cdots + (-1)^9 + (-1)^{10}$.

(The dots $\cdots$ mean that there are 21 numbers being added, one for each integer from $-10$ to 10.)
Answer: Recall that $(-a)^n= a^n$ when $n$ is an even integer and $-a^n$ when $n$ is an odd integer. In particular, when $a=1$ this identity tells us that $(-1)^n = 1$ when $n$ is even and $-1$ when $n$ is odd. This includes $(-1)^0 = 1.$  So the sum in question becomes \[
1 + (-1) + 1 + (-1) + \cdots + 1 + (-1) + 1.
\]Since $1 + (-1)=0$, we can add these numbers up in pairs to get  \[
\underbrace{1 + (-1)}_0 + \underbrace{1 + (-1)}_0 + \cdots + \underbrace{1 + (-1)}_0 + 1 = 0 + 0 + \cdots + 0 + 1 = \boxed{1}.
\]